Given a binary tree, find the maximum path sum.
For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path does not need to go through the root.
For example: Given the below binary tree,
1
/ \
2 3
Return 6
.
求二叉树上的最大路径和。
记忆化搜索,返回左右子树的最大值,
可能的最大值为左右子树返回值的和加上当前结点的值。
如果这个点的返回值为负,则完全没有必要走这条路,因此返回0.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int dfs(TreeNode* root,int &res)
{
if(!root) return 0;
int left=dfs(root->left,res);
int right=dfs(root->right,res);
res=max(res,left+right+root->val);
return root->val+max(left,right)>0?root->val+max(left,right):0;
}
int maxPathSum(TreeNode* root) {
int res=INT_MIN;
dfs(root,res);
return res;
}
};